Concept: (1+x)n=C0+x⋅C1+x2⋅C2+......+xnCn Calculation: Let S=C1+C2+C3+−−−−−+Cn =nC1+nC2+nC3+−−−−+nCn =1+nC1+nC2+nC3+−−−−+nCn−1 =nC0+nC1+nC2+nC3+−−−−+nCn−1 As we know, (1+x)n=C0+x⋅C1+x2⋅C2+......+xnCn Put x=1, So, nC0+nC1+nC2+nC3+....+nCn=2n From equation (1), S=nC0+nC1+nC2+nC3+−−−−+nCn−1 =2n−1